How many Gallons of a 12% indicator solution must be mixed with a20% indecator solution...
przesypkai4
Answered
2022-07-27
How many Gallons of a 12% indicator solution must be mixed with a20% indecator solution to get 10 gallons of a 14% solution.
Answer & Explanation
Dominique Ferrell
Expert
2022-07-28Added 18 answers
This can be solved using the following system of equations,which can be solved by a variety of methods, let me know if youwant me to solve it. .12x+.2y=.14 x+y=10 Where x is the number of gallons of the first solution, and yis the number of gallons of the second.
Israel Hale
Expert
2022-07-29Added 5 answers
Let x= the number of gallons at 20% y= the number of gallons at12% The total number of gallons: x+y = 10 gallons. The total added mixture to the solution is 0.20x +.12y = .14(10) Elminating the decimals and simplifying we have 20x + 12y = 140 Now we want to find how many gallons we need at 12%. So fromthe total number of gallons, we want to eliminate x in the equation by substituting for x=10-yin order to solve for y. Hence, 20(10-y) + 12y= 140 200 -20y +12y = 140 -8y = -60 y = 7.5 gallons Therefore, you will need to add 7.5 gallons at 12%.Hence, if you wanted to find out how many gallons at 20% you simplysolve for x in the equation x+y=10 which yields x=2.5gallons. Checking your answer, we simply substitute the values into theequation 20x +12y= 140. Then 20(2.5) + 12(7.5) = 140 140 = 140 is true